A proton enters a magnetic field of flux density 1.5 T with a velocity of `2 xx 10^(7)ms^(-1)` at an angle of 30° with the field. The force on the proton is `[e_(p)=1.6xx10^(-19)C]`

Options

A. `2.4xx10^(-12)N`
B. `2.4xx10^(-11)N`
C. `3xx10^(-5)N`
D. `3xx10^(-4)N`

Answer

To find the magnetic force on a proton entering a magnetic field, we can use the formula for the magnetic force acting on a charged particle:

\[ F_B = q v B \sin \theta \]

Where:
- \( F_B \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic flux density,
- \( \theta \) is the angle between the velocity and the magnetic field.

### Step-by-step Solution:

1. **Identify the given values:**
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Velocity of the proton, \( v = 2 \times 10^{7} \, \text{m/s} \)
- Magnetic flux density, \( B = 1.5 \, \text{T} \)
- Angle, \( \theta = 30^\circ \)

2. **Substitute the known values into the formula:**
\[ F_B = (1.6 \times 10^{-19}) \times (2 \times 10^{7}) \times (1.5) \times \sin(30^\circ) \]

3. **Calculate \( \sin(30^\circ) \):**
\[ \sin(30^\circ) = \frac{1}{2} \]

4. **Substitute \( \sin(30^\circ) \) into the equation:**
\[ F_B = (1.6 \times 10^{-19}) \times (2 \times 10^{7}) \times (1.5) \times \left(\frac{1}{2}\right) \]

5. **Simplify the expression:**
\[ F_B = (1.6 \times 10^{-19}) \times (2 \times 10^{7}) \times (0.75) \]

6. **Calculate the product:**
- First, calculate \( 1.6 \times 2 = 3.2 \)
- Then, \( 3.2 \times 0.75 = 2.4 \)
- Finally, combine with \( 10^{-19} \times 10^{7} = 10^{-12} \)

Thus,
\[ F_B = 2.4 \times 10^{-12} \, \text{N} \]

### Final Answer:
The magnetic force on the proton is \( F_B = 2.4 \times 10^{-12} \, \text{N} \).

---